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思维导图学习笔记

自己参考BoBo老师课程讲解实现：

# -*- coding: utf-8 -*-import numpy as npfrom metrics import r2_scoreclass LinearRegression(object):    def __int__(self):        self.coef_ = None  # 表示系数        self.intercept_ = None  # 表示截距        self._theta = None  # 过程计算值，不需要暴露给外面    def fit(self, X_train, y_train):        """根据训练数据集X_train, y_train训练Linear Regression模型"""        assert X_train is not None and y_train is not None, "训练集X和Y不能为空"        assert X_train.shape[0] == y_train.shape[0], "训练集X和Y的样本数要相等"        # np.linalg.inv(X) 表示求X的逆矩阵        # 不能忘了X要增加一列，第一列数据为0        ones = np.ones(shape=(len(X_train), 1))        X_train = np.hstack((ones, X_train))        self._theta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)        self.intercept_ = self._theta[0]        self.coef_ = self._theta[1:]    def _predict(self, X):        return X.dot(self.coef_.T) + self.intercept_    def predict1(self, X_test):        """给定待预测数据集X_test，返回表示X_test的结果向量"""        assert X_test.shape[1] == self.coef_.shape[0], '测试集X的特征值个数不对'        return np.array([self._predict(X) for X in X_test])    def predict(self, X_test):        """给定待预测数据集X_test，返回表示X_test的结果向量"""        assert X_test.shape[1] == self.coef_.shape[0], '测试集X的特征值个数不对'        ones = np.ones(shape=(len(X_test), 1))        X_test = np.hstack((ones, X_test))        return X_test.dot(self._theta)    def scores(self, X_test, y_test):        """根据测试数据集 X_test 和 y_test 确定当前模型的准确度"""        assert X_test.shape[0] == y_test.shape[0], '测试集X和Y的个数不相等'        return r2_score(y_test, self.predict1(X_test))

测试脚本：

# -*- encoding: utf-8 -*-from sklearn.datasets import load_bostonfrom model_selection import train_test_splitfrom linearregression import LinearRegressionboston = load_boston()X = boston.datay = boston.targetX = X[y < 50]y = y[y < 50]X_train, X_test, y_train, y_test = train_test_split(X, y, seed=666)lrg = LinearRegression()lrg.fit(X_train, y_train)# 为什么求出来的theta.shape== (13L,)print lrg.coef_print lrg.intercept_print (lrg.scores(X_test, y_test))

运行结果：

[-1.18919477e-01  3.63991462e-02 -3.56494193e-02  5.66737830e-02

 -1.16195486e+01  3.42022185e+00 -2.31470282e-02 -1.19509560e+00

  2.59339091e-01 -1.40112724e-02 -8.36521175e-01  7.92283639e-03

 -3.81966137e-01]

34.16143549624022

0.8129802602658537

实现总结：

1、上面的准确率是81.29%

2、在实现过程中，fit方法里面，忘了在X_train训练集中添加1的列向量，导致计算出来的系数矩阵参数不对

3、在pridect中，有两种实现方法，可以直接利用中间计算出来的theta值，也可以用X_test.dot(系数矩阵) + 截距举证

4、 np.hstack这个方法，参数值一个tuple，同时要注意相互之间的顺序，越是前面的矩阵则在合并后矩阵越靠前

5、在numpy中实现一个矩阵的逆矩阵用到的方法是np.linalg.inv方法

sklearn中的回归实现

# -*- encoding: utf-8 -*-from sklearn.datasets import load_bostonfrom sklearn.neighbors import KNeighborsRegressorfrom sklearn.linear_model import LinearRegressionfrom sklearn.model_selection import train_test_split, GridSearchCVdef test_regression():    boston = load_boston()    X = boston.data    y = boston.target    X = X[y < 50]    y = y[y < 50]    X_train, X_test, y_train, y_test = train_test_split(X, y)    # sklearn中的利用LinearRegression    lrg = LinearRegression()    lrg.fit(X_train, y_train)    print ('LinearRegression:', lrg.score(X_test, y_test))    # 利用KNN回归    knnrg = KNeighborsRegressor()    # 利用网格搜索找到最好的KNN参数    param_grid = [        {            'weights': ['uniform'],            'n_neighbors': [i for i in range(1, 11)]        },        {            'weights': ['distance'],            'n_neighbors': [i for i in range(1, 11)],            'p': [i for i in range(1, 6)]        }    ]    grid_search = GridSearchCV(knnrg, param_grid, n_jobs=-1, verbose=1)    grid_search.fit(X_train, y_train)    print 'GridSearchCV:', grid_search.best_estimator_.score(X_test, y_test)    print 'GridSearchCV Best Params:',  grid_search.best_params_if __name__ == '__main__':    test_regression()

运行结果：

LinearRegression: 0.7068690903842936

Fitting 3 folds for each of 60 candidates, totalling 180 fits

[Parallel(n_jobs=-1)]: Done 180 out of 180 | elapsed:    1.3s finished

GridSearchCV: 0.6994655719704079

GridSearchCV Best Params: {'n_neighbors': 10, 'weights': 'distance', 'p': 1}

sklearn中可以看到出来，线性回归还是要比KNN回归要好。此外也要注意，在使用网格搜索的时候，不能有grid_search.score（）方法，而是用grid_search.best_estimator_.score方法，因为两个实现不一样。

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